What is the integral from 0 to ln(4) of e^x-e^{-2x} ?

The integral from 0 to ln(4) of e^x-e^{-2x} is 81/32

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integral from 0 to ln(4) of e^x-e^{-2x}

\int_{0}^{l n (4)} e^{x} - e^{- 2 x} d x

\frac{81}{32}

Decimal Notation

2.53125

Solution steps

\int_{0}^{l n (4)} e^{x} - e^{- 2 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{l n (4)} e^{x} d x - \int_{0}^{l n (4)} e^{- 2 x} d x

\int_{0}^{l n (4)} e^{x} d x = 3

\int_{0}^{l n (4)} e^{- 2 x} d x = \frac{15}{32}

= 3 - \frac{15}{32}

Simplify

3 - \frac{15}{32} : \frac{81}{32}

= \frac{81}{32}

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What is the integral from 0 to ln(4) of e^x-e^{-2x} ?
The integral from 0 to ln(4) of e^x-e^{-2x} is 81/32